Question: If $a$, $b$, and $c$ are positive integers such that $\gcd(a,b) = 168$ and $\gcd(a,c) = 693$, then what is the smallest possible value of $\gcd(b,c)$?
Solution: Note that $\gcd(168,693) = 21$.  Since $\gcd(a,b) = 168 = 8 \cdot 21$, both $a$ and $b$ are divisible by 21.  Since $\gcd(a,c) = 693 = 21 \cdot 33$, both $a$ and $c$ are divisible by 21.  Therefore, $\gcd(b,c)$ must be at least 21.

If we take $a = 5544$ (which is $21 \cdot 8 \cdot 33$), $b = 168$, and $c = 693$, then $\gcd(a,b) = \gcd(5544,168) = 168$, $\gcd(a,c) = \gcd(5544,693) = 693$, and $\gcd(b,c) = \gcd(168,693) = 21$, which shows that the value of 21 is attainable.  Therefore, the smallest possible value of $\gcd(b,c)$ is $\boxed{21}$.